Ta có:

B=\(\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)  = \(\frac{3^{10}.16}{3^9.2^4}\)= \(\frac{3^9.3.16}{3^9.16}\)= 3

C=\(\frac{2^{10}.\left(13+65\right)}{2^8.104}\) =\(\frac{2^{10}.78}{2^8.104}\)  = \(\frac{2^8.2^2.78}{2^8.104}\)=  \(\frac{4.78}{104}\)  = \(\frac{4.78}{4.26}\)=\(\frac{78}{26}\)=3

=>B=C

a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)

b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)

c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{4.78}{104}=\frac{312}{104}=3\)

~ Chúc em học tốt ~

a) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\) 

\(=\frac{2^{10}.\left(13.4+65.4\right)}{2^{10}.104}+\frac{3^9.\left(3.11+3.5\right)}{3^9.16}\)

\(=\frac{312}{104}+\frac{48}{16}\)

=3+3=6

b) \(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(=\frac{1.5.6\left(1+2.2.2+4.4.4+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+9.9.9\right)}\)

\(=\frac{1.5.6}{1.3.5}\)

\(=2\)

c) 1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013

Nhận xét:Giá trị tuyệt đối của hai số liền nhau hơn kém nhau 1 đơn vị

=> Tổng trên có 2013-1+1=2013(Số hạng)

Nhóm 4 số vào một nhóm, ta được 2013:4=503 nhóm (thừa 1 số)

=>1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013

=1+(2-3-4+5)+(6-7-8+9)+...+(2010-2011-2012+2013)

=1+0+0+...+0 (có 503 số 0)

=1+0.503

=1+0

=1 

b, \(3737.43-4343.37=\left(37.101\right).43-\left(43.101\right).37=0\)

suy ra B = 0

c, \(D=\frac{2^{12}\left(13+65\right)}{2^{10}.104}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.16}{3^9.2^4}\)

\(=\frac{2^{12}.2.39}{2^{10}.2^3.13}+\frac{3^{10}.2^4}{3^9.2^4}=\frac{39}{13}+3=6\)

Gợi ý 

bn thực hiện phép tính tử mẫu bình thường , khi ra nhưng số trùng nhau bn gạch ra nháp cho đến nhưng số tối giản là ra nha 

chúc bn

học tốt

A = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

  = \(\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)

 = \(\frac{3^{10}.16}{3^9.2^4}\)

 = \(\frac{3^{10}.2^4}{3^9.2^4}=3\)

B = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

  = \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)

  = \(\frac{2^{10}.78}{2^8.104}\)

  = \(\frac{2^{10}.13.2.3}{2^8.2^3.13}\)

 = \(\frac{2^{11}.13.3}{2^{11}.13}=3\)

\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)

\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )

Để Thiên giải câu 3 cho:

(125³.7⁵ -175⁵;5):2001²⁰⁰¹

\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)

\(=\left(5^9.7^5-175:5\right):2001^{2002}\)

\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)

\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)

\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)

\(=\left(175^4.35-175^4.35\right):2001^{2002}\)

\(=0:2001^{2002}\)

\(=0\)